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Peano Axioms to Present a Rigorous Introduction to the Natural Numbers Would Take Us Too Far Afield

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Peano Axioms to Present a Rigorous Introduction to the Natural Numbers Would Take Us Too Far Afield Peano Axioms To present a rigorous introduction to the natural numbers would take us too far afield. We will however, give a short introduction to one axiomatic approach that yields a system that is quite like the numbers that we use daily to count and pay bills. We will consider a set, N,tobecalledthenatural numbers, that has one primitive operation that of successor which is a function from N to N. We will think of this function as being given by Sx x 1(this will be our definition of addition by 1). We will only assume the following rules. 1.IfSx Sy then x y (we say that S is one to one). 2. There exists exactly one element, denoted 1,inN that is not of the form Sx for any x N. 3.IfX N is such that 1 X and furthermore, if x X then Sx X, then X N. These rules comprise the Peano Axioms for the natural numbers. Rule 3. is usually called the principle of mathematical induction As above, we write x 1 for Sx.We assert that the set of elements that are successors of successors consists of all elements of N except for 1 and S1 1 1. We will now prove this assertion. If z SSx then since 1 Su and S is one to one (rule 1.) z S1 and z 1.If z N 1,S1 then z Su by 2. and u 1 by 1. Thus u Sv by 2. Hence, z SSv. This small set of rules (axioms) allow for a very powerful system which seems to have all of the properties of our intuitive notion of natural numbers. We will now give evidence for this claim. We first show that we can use the successor function to inductively define addition. We define (as indicated above) x 1 to be Sx for all x N.Ifx N is fixed (but arbitrary) and y N is such that x y has been defined then define x Sy to be Sx y. We first prove that this defines an operartion on every pair of natural numbers. Let X be the set of all y N such that x y has been defined for all x N. Then 1 X and if y X then Sy X by definition. Thus 3. implies that X N. Let us look at what this means in examples before we derive additional properties of this addition. We started with x 1 Sx for all x N. We have not yet defined any other addition for all x N. Thus we must set x S1 Sx 1 SSx for all x N. Similarly, we must define x SS1 Sx S1 SSx 1 SSSx.This continues ad infinitum. We now show that this inductive definition of addition satisfies the basic rules of arithmetic. Theorem If x,y,z N then x y z x y z (associative rule for addition). If x,y N then x y y x (commutative rule for addition). Proof. We first prove the associative rule. Let Z be the set of all z N such that x y z x y z for all x,y N. We first show that 1 Z.Infact, x y 1 Sx y x Sy (this is our definition). Now x Sy x y 1.We have thus shown that x y 1 x y 1. Thus, 1 Z. Assume that z Z we will now show that Sz Z. Now we apply the definition of addition several times x y Sz Sx y z Sx y z x Sy z x y Sz. Thus Sz Z. This implies that Z N (rule 3.). We will next prove the commutative rule. We first show that 1 x x 1 for all x N. Let X be the set of all x N such that 1 x x 1. It is clear that 1 X.Now assume that x X. Then Sx 1 SSx Sx 1 S1 x (since x X). Now, S1 x 1 Sx by definition. This implies that if x X then Sx X. Thus X N by 3. At this point we have proved that x 1 1 x for all x N. Let Y denote the set of all y such that x y y x for all x N. Then we have just shown that 1 Y. We assume that y Y. We must show that Sy Y.Toprovethis we note that if x N then x Sy Sx y Sy x (y Y). Now Sy x y Sx y x 1 y x 1 (the associative rule). Now x y 1 1 y x (1 Y). Continuing, 1 y x 1 y x y 1 x Sy x. Putting this all together we have x Sy Sy x. Thus if y Y then Sy Y. Hence Y N by rule 3. Hence the commutative rule has been proved. The next step is to define multiplication. We wish to think of it as repeated addition. We first define 1 x x for all x N.Ifu x has been defined then we set Sux u x x. Thus 1 1x x x. Theorem 1. If x,y,z N then x yz x y z (associative rule for multiplication). 2. If x,y N then x y y x (commutative rule for multiplication). 3. If x,y,z N then x y z x y x z (the distributive rule for multiplication over addition). Proof. We will prove part 3. of the theorem first by using (what else?) the principle of mathematical induction. Let X be the set of all x N such that x y z x y x z for all y,z N. Then 1 X. Assume that x X.Wemust show that Sx X. As before we will use a chain of equalities. Sxy z x y z y z by the definition. Since x X, x y z x y x z. Thus Sxy z x y x z y z. If we now apply the associative law for addition we have x y x z y z x y x z y z. Inside the first set of braces we apply the commutative rule for addition and have x y x z y z y x y x z z y x y x z z by the associative law of addition applied inside the parentheses. We now apply the commutative law of addition within the parentheses and the associative law of addition to find y x y x z z x y y x z z. By definition the last expression is Sxy Sxz. Thus Z N and the distributive law is proved. We now prove the commutative rule for multiplication by induction. Let Y y N;y x x y for all x N. We first show that 1 Y. That is to say we must show that 1 x x 1 for all x N.Since1 x x for all x N, we must show that x 1 x for all x N. We show this by showing that the set W of elements x N such that x 1 x is equal to N. We first observe that 1 W. Assume that x W then we must show that Sx W.NowSx1 x 1 1 by the definition. Since x W, x 1 x. Thus x 1 1 x 1 Sx. Thus W N. This implies that 1 Y.Now assume that y Y we must show that Sy Y.NowSyx y x x x y x. Also x Sy x y 1 x y x 1 x y x. Thus Syx x Sy. Last we prove the associative rule. Let Z be the set of all x N such that x y z x yz for all y,z N. Then 1 Z.Since1 y z y z and 1 yz y z. Assume that x Z. Then Sxy z x y z y z x yz y z z x y z y z x y y z Sxy Sxyz. This completes the proof. We will usually write xy x y. We note that there is a specific structure that we have followed in our proofs. We have an assertion for every natural number. If the assertion is true for 1 and whenever it is true for x it is true for x 1 then it is true for all natural numbers. Let us give an example of this approach. Theorem If x,y,z N and x y x z then y z. (The cancellation rule for addition.) Proof. The assertion for x is that if y,z N and if x y x z then y z.Ifx 1 then the assertion says that if 1 y 1 z then y z. The commutative rule says that we may formulate this as if y 1 z 1 then y z. This says that if Sy Sz then y z. This is true since it is rule 1. above. Now suppose that we have shown, for x, that if x y x z then y z. Suppose that Sx y Sx z. Then y Sx z Sx. Thus Sy x Sz x. Hence y x z x by rule 1. But then x y x z. Hence y z. The theorem is therefore true. The associative rule for addition allows us to introduce an order on N.Ifx,y N then we say that x y (or y x) if there exists u N such that x y u.
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